【leetcode】Monotone Increasing Digits

Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits.

(Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.)

Example 1:
Input: N = 10
Output: 9
Example 2:
Input: N = 1234
Output: 1234
Example 3:
Input: N = 332
Output: 299
Note: N is an integer in the range [0, 10^9].

 

解题思路：本题要求的是找出一个前一位不大于后一位的整数，并且这个整数不能大于输入的数。最直接的方法，是从输入数开始判断，依次减一，直到找到符合条件的数为止。更为简便的方法可以这样，从输入数的最低位开始依次和前一位比较，如果前一位大于后一位，那么前一位减1，后面所有位的值都置为9，直到遍历到最高位为止。例如输入数为762543。从最低位3开始找，次低位的是4小于5，把4和3都置为9，5减1变成4。第一次变换后的值为762499，继续找到6小于2，所以变换成759999，最后5小于7，变成699999。

代码如下：

class Solution(object):
    def monotoneIncreasingDigits(self, N):
        """
        :type N: int
        :rtype: int
        """
        sn = str(N)
        sn = sn[::-1]
        l = []
        for i in sn:
            l.append(int(i))
        for i in range(len(l)-1):
            if l[i] >= l[i+1]:
                continue
            else:
                #l[i] = 9
                for j in range(i+1):
                    l[j] = 9
                l[i + 1] -= 1
        res = 0
        count = 0
        for i in l:
            res += i*pow(10,count)
            count += 1

        return res